# Dropping an anchor from a boat

A yacht on a lake drops its anchor overboard. What happens to the water level in the lake?

1. It rises very slightly.
2. It falls very slightly.
3. It stays exactly the same.
4. It's impossible to say.

My understanding is that due to Archimedes principle, when the anchor is in the boat, it contributes to the mass of the boat, and thus the mass of water displaced. When it is thrown overboard, it is now the volume of the anchor which contributes to the amount of water displaced. Therefore, without knowing the density, mass and volume of the anchor, it is impossible to determine the effect on the water level, therefore the correct answer is option 4.

Can anyone tell me if my reasoning is correct as there is no mark scheme for the test where this question came from.

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Possible duplicate: physics.stackexchange.com/q/30268/2451 – Qmechanic Oct 30 '12 at 15:52
Another duplicate that was closed as a duplicate: physics.stackexchange.com/questions/41672/… However, I should say that from a teacher's perspective this question is somewhat unique due to the wording of "anchor". Exactly how the properties of the object is formalized creates slight differences in how the problem is approached, as shown in those two previous questions. – Alan Rominger Oct 30 '12 at 16:32
"without knowing the density, mass and volume of the anchor," Actually, it is enough to know the density (indeed enough to know if it is less than, equal to or more than that of water), and if you think about how an anchor is suppose to work you'll have enough information. – dmckee Jun 3 '14 at 3:51
If it's a sea anchor, the answer is (4); these almost float! Basically it's a drag funnel, made of sail cloth and wood, for holding direction in a storm. But for the typical anchor, which is meant to settle to the bottom, the anchor is always heavier than water, hence (2). If the anchor is suspended (as shown in the answer below), then the answer is (3). I suppose that one could build a sea anchor that actually floats, and then the answer would be (1). So one really never knows! --- the old salt. – Peter Diehr Apr 3 at 21:08

Imagine the anchor is hanging from the bottom of the boat, dangling mid-water like this:

(There is no difference between the anchor hanging in the water and sitting in the boat, since the system boat + anchor weighs the same either way.)

The water level depends on how heavy the anchor is. If you make the anchor heavier, it pulls the boat down further, pushing the water out of the way. This water goes out to the sides and raises the water level a bit. If you make the anchor lighter, the boat rises up some, leaving a gap. Water rushes in underneath the boat, and the water level goes down.

Imagine making the chain longer until at last the anchor starts to rest on the bottom of the tank. Since the bottom of the tank is supporting the anchor, it doesn't pull down on the boat as much. From the boat's perspective, it's as if the anchor got lighter. Thus, the boat rises and the water level falls.

We must assume that the anchor is more dense than water, but that is all. If you wanted to calculate how much the water level falls, you would need to know the density and weight of the anchor.

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copied from my answer to the same question on Quora:quora.com/Physics/… – Mark Eichenlaub Oct 30 '12 at 15:26
My predicted objection to this is "but the anchor is outside of the boat!", to which I would note that if you have the anchor hanging above the water, then hang it into the water, the buoyancy from the anchor decreases the anchor's weight, so the boat rises and the total displaced volume is equal. – Alan Rominger Oct 30 '12 at 16:52
Yup. Maybe another way to say it is to imagine the anchor sitting in the bottom of the boat, then sort of magically melting through the boat so it's part of the boat, then finally arriving to the outside and attaching to the outside, then being lowered by a rope. How would the water know what's going on as the anchor moves through the boat? – Mark Eichenlaub Oct 30 '12 at 17:06
Would it be different if we considered the rotation of the Earth? Since the weight of the ocean and the anchor would be different and the difference of weight varies with height (radius) I understand that this is extremely negligible, but qualitatively, would considering the Earth's rotation affect the water level? – raindrop Feb 3 '13 at 8:47

In the water the anchor displaces a volume of water equal to the volume of the anchor. In the boat, the anchor weighs the boat down which will result in the anchor effectively displacing a volume of water that would weigh as much as the anchor. since the anchor is more dense then the water it will displace more water by weight then volume. Thus, the anchor displaces more whilst in the boat.

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Because the anchor was in the Yacht before, the water level increases because of Archimedes principle. Once you put the anchor in the water the water level still rises but not as much so the overall water level drops slightly.

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Your answer is rather confusing since the anchor starts inside the boat. – LDC3 Jun 3 '14 at 3:29
While the anchor is on board, the boat was displace water equal to the anchors volume. Once the anchor is resting on the lake/sea bed it displaces water only equal to it's own volume... – dmckee Jun 3 '14 at 3:52

3 . It stays exactly the same.

because

when anchor is indside the boat, its mass will be added to the total mass of the boat and total mass will be pulled downwards.

when you throw the anchor in the water, its weight will be substracted from the boat. boat will sink less but anchor will sink the same amount minus and plus will erase each other.

no change will happen.

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The critical mistake you made is thinking that the volume of the boat without the anchor plus the volume of the anchor will equal the volume of the boat with the anchor. Mark's answer provides a helpful counterexample where the anchor starts out tied to the boat and dangling in the water. It is then clear that the submersed volume will decrease when the rope is cut. – Alan Rominger Oct 30 '12 at 16:39